452 MR. SCHUNCK ON RUBIAN AND ITS PRODUCTS OF DECOMPOSITION. 
The baryta compound, prepared by precipitating- the ammoniacal solution with 
chloride of barium, gave on analysis the following results : — 
0*2605 grm., dried at 100° C. and burnt with chromate of lead, gave 0*4640 carbonic 
acid and 0*0740 water. 
0*4055 grm. gave 0*1835 sulphate of baryta. 
In 100 parts: — 
Carbon 48*57 
Hydrogen 3*15 
Oxygen 18*60 
Baryta 29*69 
The formula C42 Hi 3 Oi3+2BaO= 2 (Ci4 H4 04+Ba0)-l-Ci4 Hs O5 requires in 100 
parts — 
Carbon 48*27 
Hydrogen 2*49 
Oxygen 1 9*93 
Baryta 29*31 
The compound with oxide of copper gave the following results : — 
0*3680 grm., dried at 100° C. and burnt with chromate of lead, gave 0*7050 carbonic 
acid and 0*1030 water. 
0*4710 grm. gave 0*1200 oxide of copper. 
These numbers correspond very nearly to the following composition : — 
Eqs. 
Calculated. 
Found. 
Carbon 
. . 14 
84 
52*50 
52*24 
Hydrogen . . . . 
. . 4 
4 
2*50 
3*10 
Oxygen 
. . 4 
32 
20*00 
19*19 
Oxide of copper . . 
. . 1 
40 
25*00 
25*47 
160 
100*00 
100*00 
Another specimen, prepared 
in exactly the 
same manner and having the same 
pearance, gave a different composition. 
0*4375 grm. gave 0*8910 carbonic acid and 0*1345 water. 
0*5530 grm. gave 0*1080 oxide of copper. 
This gives the following composition *. — 
Eqs. 
Calculated. 
Found. 
Carbon .... 
. . 56 
336 
55*17 
55*54 
Hydrogen .... 
• . 17 
17 
2*79 
3*41 
Oxygen .... 
. . 17 
136 
22*34 
21*53 
Oxide of copper . . 
. . 3 
120 
19*70 
19*52 
609 
100*00 
100*00 
The formula of this compound must be expressed 
in the following 
manner : 
3(C]4 H4 04-l-Cu0) -}-Ci4 H5 O5. 
